Linear polynomials

\(p(x)=0\cdot x+5=5\) because \(0\cdot x=0\). Thus, no \(x\)-term remains after simplification, and the polynomial has degree \(0\). A linear polynomial must have degree \(1\), so this is a constant polynomial. It is not the zero polynomial because its value is \(5\). Exam tip: remove terms with zero coefficients before finding a polynomial’s degree.

A linear polynomial has highest power 1 and is of the form \(ax+b\), where \(a\ne0\). Thus, \(7x-4\) is linear. In contrast, \(x^2+7\) has degree 2. Exam tip: check both the highest exponent and whether a variable appears in a denominator.

In (-4x+9), the coefficient of (x) is (-4) and the constant term is (9). Identify the coefficient and constant separately.

To find the zero, set the polynomial equal to zero: \(8x-16=0\). Thus, \(8x=16\), so \(x=2\). On checking, \(8(2)-16=0\); hence, \(2\) is the correct zero. Substituting \(-2\) gives \(-32\), not zero. Exam tip: the zero of a linear polynomial \(ax+b\) is \(-b/a\).

In a linear polynomial, the coefficient of x must be non-zero. On putting r=9, (r-9)x-3=(9-9)x-3=-3, which is a constant polynomial of degree 0, not a linear polynomial. For the other values, the coefficient of x is non-zero. Exam tip: In parameter-based linear polynomials, set the coefficient of x equal to zero to find when it ceases to be linear.

While adding polynomials, combine like terms. Here, \(3x+2x=5x\) and \(-7+5=-2\). Therefore, \(p(x)+q(x)=5x-2\), so option A is correct. In option C, the constant terms have been added incorrectly. Exam tip: add the \(x\)-terms and the constant terms separately.

Multiply 5 by each term inside the bracket: \(5(2x+3)=10x+15\). Then combine like terms: \(10x+15-4x=6x+15\). Therefore, \(6x+15\) is correct. \(14x+15\) would result from incorrectly adding \(4x\) instead of subtracting it. Exam tip: after opening brackets, combine only like terms.

Given \(p(x)=9-2x\). To find \(p(1)\), substitute 1 for \(x\): \(p(1)=9-2(1)=9-2=7\). Therefore, 7 is the correct answer. The number 9 is only the constant term, not the value of the polynomial at \(x=1\). Exam tip: When evaluating a polynomial, substitute the given value carefully for the variable.

In \(8x-17\), the highest power of \(x\) is \(1\), so its degree is \(1\). A polynomial of degree \(1\) is called a linear polynomial. A quadratic polynomial would have the highest power of the variable equal to \(2\), so it is not correct here. Exam tip: identify the type of a polynomial by checking the highest power of its variable.

The governing concept is the degree of a polynomial. The degree is the greatest exponent of the variable whose coefficient is not zero. A polynomial is called linear when its degree is exactly 1; it may have one term, two terms, or more terms, provided the greatest nonzero exponent is 1. For example, 3x+5 and -7x are linear polynomials, whereas x^2+1 has degree 2 and is quadratic. Thus option C correctly identifies a linear polynomial by its highest power being 1. Power 0 describes a nonzero constant polynomial, power 2 describes a quadratic polynomial, and the number of terms does not determine whether a polynomial is linear.

In \(13u+9\), the variable \(u\) can be written as \(u^1\), so its power is 1. The number 13 is the coefficient of \(u\), and 9 is the constant term; neither is the power. Hence, this is a linear polynomial in \(u\). Exam tip: When no exponent is written on a variable, its power is 1.

The general form of a linear polynomial is (ax+b) where (a\neq0). If (a=0), the degree will not remain (1).

The degree of a polynomial is the greatest exponent of the variable with a non-zero coefficient. In \(p(x)=12x+1\), the highest exponent of \(x\) is \(1\), so its degree is \(1\). The number \(12\) is only the coefficient of \(x\), not the degree. Exam tip: Look for the highest power of the variable, not its coefficient.

In the polynomial 25-9x, the term containing x is -9x. The number multiplying x is -9, so the coefficient of x is -9. The option 9 is incorrect because the negative sign is part of the coefficient. Exam tip: Always include the sign immediately before the variable when identifying a coefficient.

A constant term is a term that contains no variable. In \(x+18\), \(x\) is the variable term, while \(18\) has no variable. Therefore, the constant term is \(18\). Although 0 can be a constant, it is not the constant term in this polynomial. Exam tip: identify the term with no letter or variable to find the constant term.

The polynomial 7x+4 has an x-term and the constant term 4, but no x²-term. It can therefore be written as 0x²+7x+4, so the coefficient of x² is 0. Here, 7 is the coefficient of x, not of x². Exam tip: If a term of a given power is missing, its coefficient is 0.

In \(7x+5\), the highest power of x is 1, so it is a linear polynomial. \(4x^2-3\) has degree 2 and is quadratic. Exam tip: identify a polynomial’s degree by its highest exponent.

The given polynomial is \(p(x)=x-20\). It can be written as \(1x+(-20)\), so it is a linear polynomial of the form \(ax+b\), where \(a=1\) and \(b=-20\). The form \(ax^2+b\) is quadratic and must contain an \(x^2\) term. Exam tip: identify the type of a polynomial from its highest power; here, the highest power of \(x\) is 1.

In (0x+12), the coefficient of x is 0, so 0x=0 and the expression simplifies to 12. Since 12 is a non-zero constant, it is a constant polynomial with degree 0. A linear polynomial must have a non-zero coefficient of x. Exam tip: Remove terms with zero coefficients before finding the degree of a polynomial.

In a linear polynomial, the coefficient of \(x\) must be non-zero. Substituting \(m=8\) gives \(m-8=0\), so the expression becomes \(0x+6=6\). This is a constant polynomial, not a linear polynomial. For example, when \(m=7\), the coefficient of \(x\) is \(-1\), so it is still linear. Exam tip: a linear polynomial has highest power 1 with a non-zero coefficient of the variable.